Question

Water is a dipole. A water molecule in vacuum has a dipole moment of \(0.62 \times 10^{-29} \mathrm{C} \mathrm{m}\). (a) Assuming that this dipole moment is due to charges \(+e\) and \(-e\) at distance \(\ell\), calculate \(\ell\) in \(A\). (b) Consider the water molecule as a sphere with radius 1.5A. If the dipole moment in (a) is due to charges \(+q\) and \(-q\) at the north and south poles of the sphere, how large is \(q\) ?

Short answer

\( \ell \) = 3.875 Å, \( q\) = 2.07 \times 10^{-20} C

Step-by-step solution

Title - Relationship between dipole moment and charge distance

The dipole moment \(\mu\) is related to the charge \(e\) and distance \((\ell)\) between charges by the equation: \[\mu = e \cdot \ell\]. Given, \[\mu = 0.62 \times 10^{-29} \, \mathrm{C} \cdot \mathrm{m}\] and \[e = 1.6 \times 10^{-19} \, \mathrm{C}\].

Title - Solving for distance \(\ell\)

Rearrange the formula to solve for \(\ell\): \[\ell = \frac{\mu}{e}\]. Substitute the given values: \[\ell = \frac{0.62 \times 10^{-29}}{1.6 \times 10^{-19}} = 3.875 \times 10^{-11} \, \mathrm{m}\]. Convert this value to angstroms (1 angstrom = 10^{-10} meters): \[\ell = 3.875 \, \mathrm{A}\].

Title - Sphere radius and charge distance

Considering the water molecule as a sphere with radius \ R = 1.5 \, \mathrm{A} \, and assuming the dipole moment is due to charges \( \pm q\) at the north and south poles, the distance between these charges would be the diameter of the sphere (2R): \[\ell = 2R = 2 \cdot 1.5 \, \mathrm{A} = 3 \, \mathrm{A}\].

Title - Solving for \( q \)

Use the dipole moment formula to find \( q \): \[ \mu = q \cdot \ell \]. Rearrange to solve for \( q \): \[ q = \frac{\mu}{\ell} \]. Substitute the known values: \[ q = \frac{0.62 \times 10^{-29}}{3 \times 10^{-10}} = 2.07 \times 10^{-20} \, \mathrm{C}\].

Key concepts

dipole moment

A dipole moment is a measure of the separation of positive and negative charges in a system. It is a vector quantity, with both magnitude and direction. The dipole moment \( \mu \) is calculated using the formula: \[ \mu = q \cdot d \], where \( \mu \) is the dipole moment, \( q \) is the charge, and \( d \) is the distance between the charges. If the charges are closer together, the dipole moment will be smaller, and if they are farther apart, the dipole moment will be larger.

• The unit of dipole moment is Coulomb meter (C·m).
• In molecules, dipole moments result from differences in electronegativity between atoms.
• A dipole moment is important as it influences molecular interactions and properties like boiling point, solubility, and intermolecular forces.

Understanding dipole moments helps in studying molecular behavior and predicting how molecules will interact with each other in different environments. For instance, in water, the separation of charges creates hydrogen bonding, contributing to water's unique properties.

charge distance

The distance between charges, often denoted as \( \ell \), is crucial in calculating the dipole moment. It refers to the physical separation between a positive charge \( +q \) and a negative charge \( -q \). Using the formula \[ \mu = q \cdot \ell \], we can see how \( \ell \) directly influences the dipole moment.In the case of the water molecule, the dipole moment is given, and we need to find \( \ell \). The rearranged formula for \( \ell \) is: \[ \ell = \frac{\mu}{q} \], providing us insight into how the distance affects the overall dipole.

• To solve for \( \ell \), divide the dipole moment \( \mu \) by the charge \( q \).
• For example, with \( \mu = 0.62 \times 10^{-29} \) Cm and \( q = 1.6 \times 10^{-19} \) C:
• \[\ell = \frac{0.62 \times 10^{-29}}{1.6 \times 10^{-19}} = 3.875 \times 10^{-11} \ \mathrm{m} \]. Converting to angstroms, this yields 3.875 A.

The correct distance calculation is essential for understanding the spatial structure and polarity of molecules.

water molecule dipole

The water molecule (H2O) is a classic example of a dipole. Due to the bent shape and the difference in electronegativity between hydrogen and oxygen, water possesses a significant dipole moment. The oxygen atom is more electronegative, pulling the shared electrons closer, making it partially negative, while the hydrogens become partially positive. This separation creates a dipole moment in the water molecule.

• The measured dipole moment of water in vacuum is \( 0.62 \times 10^{-29} \ \mathrm{C} \cdot exm \).
• In a spherical model of water with a radius of 1.5 A, the dipole moment derives from charges \( +q \) and \( -q \) at opposite poles.
• Using \[ \ell = 2R = 3 \A \] (where \( 2 \cdot 1.5 A \) is the distance between the charges \( \pm q \)), the dipole moment formula becomes: \[ q = \frac{\mu}{\ell} = \frac{0.62 \times 10^{-29}}{3 \times 10^{-10}} = 2.07 \times 10^{-20} \ C \].

This calculated charge helps in understanding the distribution of charges in a water molecule. The dipole moment explains many of water's physical properties, such as high boiling point and surface tension. Recognizing how the water dipole functions provides insight into its role in supporting life and in various chemical interactions.